use stokes theorem to evaluate the line integral

To evaluate the line integral using Stokes' theorem, we need to find the curl of the vector field F and then calculate the surface integral of the curl over the surface enclosed by the given curve C.
Let's start by finding the curl of the vector field F:
$\nabla \times 𝐅=\left|\begin{array}{ccc}𝐢& 𝐣& 𝐤\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ -y& x& 0\end{array}\right|$
Expanding the determinant, we have:
$\nabla \times 𝐅=(\frac{\partial }{\partial y}(0)-\frac{\partial }{\partial z}(x))𝐢-(\frac{\partial }{\partial x}(0)-\frac{\partial }{\partial z}(-y))𝐣+(\frac{\partial }{\partial x}(x)-\frac{\partial }{\partial y}(-y))𝐤$
Simplifying further, we get:
$\nabla \times 𝐅=-𝐢-𝐣$
Now, we need to find the surface enclosed by the given curve C, which is the boundary of the ellipse:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Since z = 0, the surface lies in the xy-plane.
To evaluate the line integral using Stokes' theorem, we can convert it into a surface integral over the surface S enclosed by the curve C:
${\oint }_C𝐅·d𝐫={\iint }_S(\nabla \times 𝐅)·d𝐒$
Substituting the values we obtained earlier:
${\iint }_S(-𝐢-𝐣)·d𝐒$
Now, we need to parameterize the surface S. Since the surface lies in the xy-plane, we can use the parameterization:
$𝐫(u,v)=⟨a\cos (u),b\sin (v),0⟩$
where $0\le u\le 2\pi$ and $0\le v\le \pi$.
Next, we calculate the cross product of the partial derivatives of $𝐫$ with respect to u and v:
$\frac{\partial 𝐫}{\partial u}=⟨-a\sin (u),0,0⟩$
$\frac{\partial 𝐫}{\partial v}=⟨0,b\cos (v),0⟩$
Taking the cross product, we get:
$\frac{\partial 𝐫}{\partial u}\times \frac{\partial 𝐫}{\partial v}=⟨0,0,ab\sin (u)\cos (v)⟩$
The magnitude of this cross product is $|\frac{\partial 𝐫}{\partial u}\times \frac{\partial 𝐫}{\partial v}|=ab\sin (u)\cos (v)$.
We can set up the surface integral as follows:
${\iint }_S(-𝐢-𝐣)·d𝐒={\iint }_S(-𝐢-𝐣)·\left(\frac{\partial 𝐫}{\partial u}\times \frac{\partial 𝐫}{\partial v}\right)dudv$
Substituting the magnitude of the cross product and the parameterization of $𝐫$, we have:
${\iint }_S(-𝐢-𝐣)·(0,0,ab\sin (u)\cos (v))dudv$
Since the surface lies in the xy-plane, the z-component of the cross product is zero, so we can ignore it in the dot product:
${\iint }_S(-𝐢-𝐣)·(0,0,ab\sin (u)\cos (v))dudv$
Expanding the dot product and simplifying, we get:
${\iint }_S-ab\sin (u)\cos (v)dudv$
Now, we can determine the limits of integration for u and v. Since $0\le u\le 2\pi$ and $0\le v\le \pi$, the limits of integration are as follows:
${\int }_0^{\pi }{\int }_0^{2\pi }-ab\sin (u)\cos (v)dudv$
Integrating with respect to u first, we have:
${\int }_0^{\pi }{[-ab\cos (u)\cos (v)]}_{u=0}^{u=2\pi }dv$
Simplifying the limits of integration, we get:
${\int }_0^{\pi }-ab\cos (2\pi )\cos (v)+ab\cos (0)\cos (v)dv$
Since $\cos (2\pi )=\cos (0)=1$, the integral becomes:
${\int }_0^{\pi }-ab\cos (v)+ab\cos (v)dv$
The two terms cancel each other out, resulting in:
${\int }_0^{\pi }0dv=0$
Therefore, the line integral evaluated using Stokes' theorem is 0.